Emily is testing her baby’s bath water and finds that it is too cold, so she adds some hot water from a kettle on the stove. If Emily adds 2.00 kg of water at 80.0° C to 20.0 kg of bath water at 27.0° C, what is the final temperature of the bath water?
the hot water looses heat , and the bath water gains heat
[80.0ºC - (final temp)] * 2.00 kg = [(final temp) - 27.0ºC] * 20.0 kg
To find the final temperature of the bath water, we can use the principle of heat transfer, which states that the heat gained by the cold water is equal to the heat lost by the hot water.
Let's assume the final temperature of the bath water is T.
The formula used to calculate heat transfer is:
Q = m * c * ΔT
Where:
Q = heat transferred (in joules)
m = mass of the substance (in kg)
c = specific heat capacity (in J/kg°C)
ΔT = change in temperature (in °C)
Now, let's calculate the heat gained by the cold water and the heat lost by the hot water:
Heat gained by cold water:
Q_cold = m_cold * c_water * (T - T_cold)
Where:
m_cold = mass of cold water (20.0 kg)
c_water = specific heat capacity of water (4.18 J/g°C) (approximately)
T_cold = initial temperature of cold water (27.0°C)
We need to convert the specific heat capacity from J/g°C to J/kg°C by dividing by 1000:
c_water = 4.18 J/g°C / 1000 = 0.00418 J/kg°C
Substituting the values:
Q_cold = 20.0 kg * 0.00418 J/kg°C * (T - 27.0°C)
Heat lost by hot water:
Q_hot = m_hot * c_hot * (T - T_hot)
Where:
m_hot = mass of hot water (2.00 kg)
c_hot = specific heat capacity of water (4.18 J/g°C) (approximately)
T_hot = initial temperature of hot water (80.0°C)
Converting the specific heat capacity:
c_hot = 0.00418 J/g°C / 1000 = 0.00418 J/kg°C
Substituting the values:
Q_hot = 2.00 kg * 0.00418 J/kg°C * (T - 80.0°C)
As per the principle of heat transfer, the heat gained by the cold water is equal to the heat lost by the hot water:
20.0 kg * 0.00418 J/kg°C * (T - 27.0°C) = 2.00 kg * 0.00418 J/kg°C * (T - 80.0°C)
Now, we can solve this equation to find the value of T, which represents the final temperature of the bath water.
20.0 kg * 0.00418 J/kg°C * T - 20.0 kg * 0.00418 J/kg°C * 27.0°C = 2.00 kg * 0.00418 J/kg°C * T - 2.00 kg * 0.00418 J/kg°C * 80.0°C
(20.0 kg * 0.00418 J/kg°C - 2.00 kg * 0.00418 J/kg°C) * T = 2.00 kg * 0.00418 J/kg°C * 80.0°C - 20.0 kg * 0.00418 J/kg°C * 27.0°C
18.8 kg * 0.00418 J/kg°C * T = 2.00 kg * 0.00418 J/kg°C * 80.0°C - 20.0 kg * 0.00418 J/kg°C * 27.0°C
Simplifying further:
0.079104 * T = 0.067032 - 0.022428
0.079104 * T = 0.044604
T = 0.044604 / 0.079104
T ≈ 0.5656
Therefore, the final temperature of the bath water after Emily adds the hot water from the kettle is approximately 0.5656°C.